For the reaction
\(2NO (g) + O_2 (g) \rightarrow 2NO_2 (g)\)
the volume of the system is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to \(O_2\) and second order with respect to NO, the rate of the reaction will:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Increase to eight times of its initial value
Molar concentration is moles per unit volume, so halving the volume doubles the concentration of every gas. With the orders given, the rate law is:
\( \displaystyle r_1=k[\ce{NO}]^2[\ce{O2}] \)
After the volume is halved, [NO] and [\( \ce{O2} \)] both become twice as large:
\( \displaystyle r_2=k(2[\ce{NO}])^2(2[\ce{O2}])=8\,k[\ce{NO}]^2[\ce{O2}] \)
\( \displaystyle \frac{r_2}{r_1}=8 \)
The rate increases to eight times its initial value: a factor of 4 from the second order in NO and a factor of 2 from the first order in \( \ce{O2} \).